You have found the following ages (in years) of all 6 bears at your local zoo: $ 31,\enspace 16,\enspace 19,\enspace 13,\enspace 38,\enspace 18$ What is the average age of the bears at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{31 + 16 + 19 + 13 + 38 + 18}{{6}} = {22.5\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $31$ years $8.5$ years $72.25$ years $^2$ $16$ years $-6.5$ years $42.25$ years $^2$ $19$ years $-3.5$ years $12.25$ years $^2$ $13$ years $-9.5$ years $90.25$ years $^2$ $38$ years $15.5$ years $240.25$ years $^2$ $18$ years $-4.5$ years $20.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{72.25} + {42.25} + {12.25} + {90.25} + {240.25} + {20.25}} {{6}} $ $ {\sigma^2} = \dfrac{{477.5}}{{6}} = {79.58\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{79.58\text{ years}^2}} = {8.9\text{ years}} $ The average bear at the zoo is 22.5 years old. There is a standard deviation of 8.9 years.